\frac{3^{x^2}}{(\sqrt3)^{x+\frac{1}{2}}}>\sqrt[4]{3}
\frac{3^{x^2}}{3^{\frac{1}{2}(x+\frac{1}{2})}}>3^{\frac{1}{4}}
\frac{3^{x^2}}{3^{\frac{1}{2}x+\frac{1}{4}}}>3^{\frac{1}{4}}
3^{x^2-(\frac{1}{2}x+\frac{1}{4})}>3^{\frac{1}{4}}
3^{x^2-\frac{1}{2}x+\frac{1}{4}}>3^{\frac{1}{4}}
x^2-\frac{1}{2}x+\frac{1}{4}>\frac{1}{4}
x^2-\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}>0
x^2-\frac{1}{2}x+\frac{2}{4}>0
x^2-\frac{1}{2}x+\frac{1}{2}>0/*2
2x^2-x+1>0
a=2, b=-1, c=1
\Delta=b^2-4ac=1-4*2*1=9
\sqrt\Delta=3
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{1-3}{2*2}=-\frac{2}{4}=-\frac{1}{2}
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{1+3}{4}=1
x\in(-\infty;-\frac{1}{2})\cup (1;+\infty)